Simplify Design Method for Slender Hybrid Columns

Simplify Design Method for Slender Hybrid Columns Try MENG1, P. KEO2, Hugues SOMJA3 Abstract: In this paper, simplify method based on moment magnification approach of Eurocode for designing concrete columns was modified by adding a calibrate factor (αk) in order to apply for hybrid columns. The proposed simplify method significantly depended on effective flexural stiffness (EI)eff of the hybrid columns. The expressions of (EI)eff were introduced differently by EC2 [1] and EC4 [2], the expression of effective flexural stiffness for applying on hybrid columns was given in this paper. The correction factors (Kc) and (Ksa) which corresponding to concrete and equivalence between reinforcing rebar and structural steel were respectively concluded in the expression of flexural stiffness (EI)eff of hybrid columns. The finite element model developed by P. Keo [3] was used to run for 1080 cases of parametric study, and the results from parametric study will serve to formulate the new expression of the correction factors (Kc) and (Ksa) which based on secant modulus. The failure mode of hybride columns was divided into two types, one was failure of resistance of the cross-section, and another was the failure of instability of hybrid columns. The verification of hybrid columns was also determined, plastic resistance bending moment due to instability failure of hybrid columns was introduced in this paper by multiplying a reduction factor α! with plastic resistance bending moment of the cross-section (Mpl,N,Rd). Idea was to verify the bending moment (MEd,SM) calculated from simplify design method of hybrid columns with plastic resistance bending moment (MRd,Alpha) calculated from reduction factor α! . Key Word: hybrid column; Eurocode; simplify method; moment magnification; flexural stiffness; correction factor; secant modulus; instability failure; reduction factor.

1. Introduction In the recent decade, due to the numerous of highrise building and bridge construction, the demanding of high strength column has been increased from year to year. Classical reinforced concrete column has no longer being an option for these constructions due to the reasons of huge dimension for cross-section; time consuming in construction; not enough strength for axial load and bending moment. The idea of composite structure has been proposed and studied, in EC4 [2] the design code for single encased steel profile in reinforced concrete column has been provided. Nowadays, even the typical composite column is not a solution for certain constructions of super high-rise building where the column is generally heavily loaded. 1

Master research, Department of Civil engineering at INSA-Rennes, France and Research assistance at LGCGM, INSA-Rennes. 2 Eng., Ph.D, Structural engineering research group, LGCGM, INSA-Rennes. 3 Maître de conférence, Professor and researcher at Dpt. Civil Engineering and urban, INSA-Rennes.

International Finance Center Tower 2 (IFC2) in Hong Kong, various forms of composite column have been introduced in this project. Interestingly, megacolumns with more than one steel profile encased in reinforced concrete column ere used whilst there is no available guidance introduced for such member in Eurocode. There are three principal materials considered in hybrid columns and they are concrete; steel profiles and reinforcement. The combination of those materials has contributed to produce higher strength in compression and bending; large flexural stiffness therefore leading to increased buckling resistance. There are very few researchers involved in the research of hybrid columns. The contribution to the topic is mostly done by P.Keo and co-workers [3]. P.Keo developed a finite element model of hybrid columns in which the nonlinear geometry and material are taken into account. A parametric study on the second order effects in slender hybrid columns was then performed by using the validated FE model. The results of parametric study was served as a reference to evaluate the applicability to hybrid columns design of the simplify method given in EC2

and EC4 [2] based on the moment magnification cross-section (Mpl,N,Rd) cannot be used directly to approach. The comparison of the results given by verify the resistance of the columns, particularly in simplified method of EC2 and EC4 against the ones case instability failure occurs. The direct use of from FEA showed that there is a wide discrepancies Mpl,N,Rd in the research paper of P.Keo [3] may lead between both pairs where for half of case studies to underestimate the effective flecural stiffness of the both EC2 and EC4 method give unconservative column. Mpl,N,Rd should be replaced by the secondresults. It is worth mentioning that in the momentmoment oforder resistqnce moment of the column or the inertia (I i). Before starting to run FEA, an editing in Matlab code is nece magnification method proposed by P.keo et al. [3],means to store second-order bending obtained fromload FEA all values of stress; strainmoment corresponding to the failure and at the mos the criterion to verify the resistance of the columns issection within (MRd,2nd,EF ) at ultimate load and in the critical crossthe whole length of hybrid columns. By timing each secant modulus of i that the second-order bending moment has to beto the moment section. Due oftothat those reasons, in this we stiffness will (EI)sec,i , of inertia element and then sum paper, all flexural lesser or equal to plastic bending moment of thefinally obtain perform a parametric study to figure out the (EI)c,sec ; (EI) s,sec ; (EI)a,sec . cross-section (Mpl,N,Rd) taking into account the • Concrete: expression of correction factor (Kc; Ksa) for the σ! of concrete, reinforcement and applied normal force. However, the collapse of flexural stiffness σ! ; ε! ; I! → E!,!"#,! = → (EI)!,!"#,! = E!,!"#,! ×I! → (EI)!,!"# = (EI)!, slender hybrid columns can be caused by material or structural steel εin! hybrid columns, respectively, in moment of inertia (Ii). Before starting to run FEA, an editing in Matlab code is necessary, it • Reinforcement rebar:a better estimation of flexural stiffness. instability failure. The plastic bending moment offailure the load order to critical give means to store all values of stress; strain corresponding to the and at the most •

σ!

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(c)

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means to store all values of stress; strain corresponding to the failure load and at the most critical

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σ! ; ε! ;ofI! HSRCC1, → E!,!"#,! → (EI)!,!"#,! of= 10m, E!,!"#,!f×I!=35Mpa; → (EI)!,!"# = (EI)!,!"#,! In figure 3.2(c), an example with= the ck ε! column length fsk=500Mpa; fyk=235Mpa; φef = 1 and eo =0, the value of flexural stiffness corresponding to each 3.3. The procedure for obtaining the value of correction factors (Kc,EF; Ksa,EF) • Structural steel: element of cross-section have been put into the colour in which orange represent minimum value σ! flexural stiffness The objective to of correction factors is to calculate flexural stiffness of hybride columns w and green represent maximum In this example, σ! ; ε!value. ; I! → E!,!"#,! = the → (EI)!,!"#,! =starts E!,!"#,!reduce ×I! →from (EI)!,!"# = (EI)!,!"#,! ε column reaches its ultimate the mid point of the cross-section and this because of ! the eccentricity of axial force is equal to stage. As stated before, the correction factors will be ca zero and middle steel was started to plasticize which lead to reduce the flexural stiffness of Master thesis-MENG Try 13 structural steel.

InThe [fig. 2], the force-displacement curveflexural has been presented, after running objective of correction factors is to calculate stiffness of hybride columns when the FE model for a specific case of hybrid column reaches its ultimate stage. As the correction factors be calculated In [fig. 3], M-N interaction curve was column, ultimate axial force andstated totalbefore, bending moment werewilldetermined. Master thesis-MENG Try 13 presented, two types of bending moment were bending moment obtained from FE model (MEd,EF) and plastic (b) (a) resistance bending moment of the cross-section (M pl,N,Rd) and vary with different eccentricities. For the same axial

3. The procedure for calculating numerical value of (Kc,EF; Ksa,EF) 3.1. of bending moment magnification 3.3. Method The procedure for obtaining the value of correction factors (Kc,EF; Ksa,EF)

(c)

moment of inertia (Ii). Before starting to run FEA, an editing in Matlab code is necessary, it

2.492870

Figure 1 finally obtain (EI)c,sec ; (EI)s,sec ; (EI)a,sec . In figure 3.2(c), an example of HSRCC1, with the column length of 10m, fck =

fsk=500Mpa; fyk=235Mpa; φef = 1 and eo =0, the value of flexural stiffness corresponding element of cross-section have been put into the colour in which orange represent minimu (c) σ! and green represent maximum value. In this example, the flexural stiffness starts to redu σ! ; ε! ; I! → E!,!"#,! = → (EI) (EI) !,!"#,! = E!,!"#,! ×I! → (EI)!,!"# = !,!"#,! ε! the mid point of the cross-section and this because of the eccentricity of axial force is zero and middle steel was started to plasticize which lead to reduce the flexural stif • Reinforcement rebar: Figure 3.2: Fibre cross section of concrete and steel structural steel. σ!

Concrete:

1.70699 ####### 1.65344 #######

section within the whole length of hybrid columns. By timing each secant modulus of i element Figure 3.2: Fibre cross a b section of concrete and steel to the moment of inertia of that element and then sum all flexural stiffness (EI)sec,i , we can (b)

(a)

(EI)!,!

(b)

; ε! ; I! → of E!,!"#,! = → (EI)!,!"#,! = E!,!"#,!by ×I!strain → (EI) (EI)value !,!"#,! of stressor strain. In finite element model, Secantσ!modulus elasticity divided at !,!"# any=given ε! is stress the cross-section ofrebar: hybrid columns were divided into multi-fibres as show in [Fig. 1(a)], each element of the cross• Reinforcement section contains differentσ! stress (σi); strain (εi) and second moment of inertia (Ii). Before starting to run FE model, σ! ; ε! ; I! → E!,!"#,! = → (EI)!,!"#,! = E!,!"#,! ×I! → (EI)!,!"# = (EI)!,!"#,! an editing in Matlab codeε! is necessary, it means to store all values of stress; strain of each element corresponding to • Structural steel: the failure load and at the most critical section within the whole length of hybrid columns. A typical cross-section σ! σ! ; ε! ;columns, I! → E!,!"#,! → (EI)value E!,!"#,! ×I! →stiffness (EI)!,!"# of = each (EI)element !,!"#,! =of !,!"#,! of hybrid the=numerical flexural (i) was put into color in [Fig. 1(b)]. ε! σ!

σ! → (EI)!,!"#,! = E!,!"#,! ×I! → (EI)!,!"# = ε! σ! ; ε! ; I! → E!,!"#,! = Concrete: •

•

Structural steel:

(EI)!,! obtain (EI)c,sec ; (EI)s,secfor ; (EI) a,sec . 2.finally The procedure obtaining secant modulus form FEA

→ (EI)!,!"#,! = E!,!"#,! ×I! → (EI)!,!"# = σ! modulus ; ε! ; I! → E!,!"#,! = section within the whole length of hybrid columns. By timing each secant of i element ε! to the moment of inertia of that element and then sum all flexural stiffness (EI)sec,i , we can

(a)

!" the ultimate bending moment (M!",!! ) that we obtained from FEA is not always equal to plastic

bending moment resistance (M!",!,!" ). In [figure 3.4], M-N interaction curve was presented, blue curve represents bending moment resistance due to the axial force and red curve was bending load, ultimate bending moment (MEd,EF calculated by FE was smaller than MWhen graphs was proved pl,N,Rd, both moment obtained from FEA) and vary with different eccentricities. eccentricity a the failure mode of hybrid andthe thecolumn incorrect of M directly moment magnification smallcolumns value, then wasuse failed bypl,N,Rd instability andinthis is because of the ratiomethod Ncr/NEdin = the research paper of P.Keo 2.8 [3]. and the columns started to buckle. In contrast, as the eccentricity increase and failure of 𝑀!",!" hybrid columns was caused from𝑀cross-section resistance meanwhile, axial force (NEd) also !",!,!" 𝐾!" = → 𝐾!" = reduced and ratio Ncr/NEd =7.3. 𝑀!",! 𝑀!",! In general, there are two types of failure mode for hybrid columns:

•

Fu is the ultimate load cause by failure of section resistance !" M!",!! ≈ M!",!,!"

•

Fu is the ultimate load cause by instability failure !" M!",!! < M!",!,!"

The result of this problem is due to the reason that for the same cross-section; material strength but with different length of hybrid columns. The failure mode of longer columns could be highly caused by instability failure compare to the short columns, which has the failure point caused by resistance of the cross-section. This is to show that it not always true for the expression K !" =

!!",!,!" !!",!

!" in the previous study, then Mpl,N,Rd should replace by M!",!! .

!" M!",!! k = Figure 2 Figure 3.3: Force-displacement!" M!",!

Figure 3 Figure 3.4: Interaction curve

Using K anc, example from [section 3.2], with the same strength of material for all 3.2. Calculation Ksa from flexural stiffness 3.3.2. Calculate Kcof ; secant KHSRCC1 secant flexural stiffness s; Ka from

In reality, the correction factors be considered the reducing factor for stiffness. (EI) cross-sections but varycan eccentricity (0;as0.02; 0.05; 0.3; 0.5; 1;flexural 2;for 4; 6), in [figure 3.3],sec, which we In reality, the correction factors canfrom be considered as the0.1; reducing factor flexural stiffness. obtain from(EI) finite element analysis took account of theofcreep caused plasticity of steel, wetoobtain fromalready finite provide element analysis already took account ofby theconcrete, creep caused by sec which corresponding e=0, FEA could us the diagram ultimate load and displacement. In eccentricity,concrete, etc. example Dividing (EI)ultimate with axial the gross weDividing finally obtained numerical value correction factors plasticity of eccentricity, (EI)sec the gross (EI), weoffinally sec steel, this the force(EI), ofetc. hybrid column waswith Fthe u = 6837 KN with MEd, EF numerical valuesteps of correction factors (Kc; Ks; Ka) and the detail calculation steps (Kc,EF, Ksa,EFobtained ) and thethe detail calculation were presented: =441KN.m < Mpl,N,Rd = 572 KN.m as shown in [figure 3.4]. Both graphs above also proved the were presented below: incorrect use of MPl,N,Rd directly in moment magnification method in the previous research of P. FEA Keo et al. [3] that we have shown the calculation step in [section 2.3.3]. (!")!,!"# ; (!")!,!"# ; (!")!,!"# Master thesis-MENG Try

14 (EI) (EI)!"",!! = (EI)!,!"# + !,!"# + (EI)!,!"#

(EI)!"" =

(EI)!,!"# (EI)!,!"# (EI)!,!"# . E!" I! + . E! I! + . E! I! E!" I! E! I! E! I!

k !,!" =

(!")!,!"# !!" !!

; k !,!" =

K !" =

1−

π! l!

(!")!,!"# ! ! !!

β N!"

; k !,!" =

(!")!,!"# ! ! !!

> 1.1

. (EI)!"",!!

compare K !" and K !"

The numerical value of Kc,EF and Ksa,EF will be used to draft the graphs both in a specific case !" with the same Moreover, in order to obtain the ultimate bending moment from finite element analysis (M!",!! ), material strength; geometry and another graph presented all data of Kc,EF and Ksa,EF respectively versus parameters FEA code in Matlab will be needed to edit related to slenderness ratio; creep coefficient; material strength and axial load. By noted into each case of parametric study, the parameters that are mostly influence the results of Kc,EF and Ksa,EF were found out. Finally, those parameters will be taken part into the formulation 15 fo Kc, Ksa and additional constant will be also considered. Master thesis-MENG Try

250

250

expressions. In this report, there are two cross-sections were selected to study. HSRCC1 [figure 3.1] with the dimension of 800x250 was fully encased with 3 steel profiles of HEB120, HSRCC5 with the dimension of 2000x2000 was also fully encased with 4 steel profiles of 4. HD400x1086 Case of parametric study corner. For HSRCC5 due to reason of bi-symmetry, in finite at four different 60 60 60 60 260 260 HSRCC1 and 260 HSRCC5 [fig. 4], There are two types of cross-section have260 been selected for studying and they are: element analysis, only half of cross-section will be modelled. Both sections will share the same 800 study. For both cross-section of HSRCC1 and800HSRCC5, three different there are 1080 cases of parametric concrete strength which vary from 35Mpa; 60Mpa and 90Mpa and steel strength started from strengths of concrete and steel which vary from (35; 60; 90)Mpa and (235; 355; 460)Mpa were adopted (a) Cross-section Cross-section 235Mpa; 355Mpa and 460Mpa,HSRCC1. for reinforcement rebar is (b) 500Mpa. Beside that,HSRCC2. long-term effect respectively. Moreover, for better understanding the slenderness of hybrid columns, different types of column suchwere as creep 0 toaccount 1, consider with different ineccentricity and length of length used incoefficient parametric will study.change In orderfrom to take of creep phenomenon concrete, two coefficient in selected. [table 3.1], we obtained 540 casesonly forhalf firstofsection and another 540 cases for in φef as (0;show 1) were Duefinally to the reason of bi-symmetry, cross-section of HSRCC5 was model 6.3 Parametric study and assessment of simplified methods of EC2 theelement secondanalysis one. (FEA). finite 500

85

45

45

40

2000

1000

45

45

500

3000

500

1000

and EC4

85

40

500

45

45

500

500

250

250

based on EC2 also include in each case. In [table 3.1] and [table 3.2], the detail of cross-section and position of structural steel; reinforcing rebar were presented and detail of structural steel was also shown in [figure 3.5]. Due to the reason of bi-symmetry, only half of cross-section of 60 60 60 60 260 260 260 260 HSRCC5 was model in FEA and steel ratio was divided by two. One thing to be noted, the steel 40 40 85 85 800 orientation both is different, in800‘H’500and HSRCC5 ‘I’ for the 500 was 500 put500 500 for HSRCC1 1000 sections1000 500 first section steel profile 2000 3000 second section. (a) Cross-section HSRCC1. (b) Cross-section HSRCC2. (a) HSRCC1

(b) HSRCC5

31

3.2.1: Different The procedure for obtaining Table cases of cross-section

500

85

Figure 4: Cross-section(d) Cross-section HSRCC5. (c) Cross-section HSRCC3. Figure 3.1: Detail cross section for hybrid column

secant modulus from FEA 1000

250

Secant modulus of elasticity is stress divided strain at any given value offs (Mpa) stress or strain. In section Length (m) fck (Mpa) φef by Eccentricity (m) fa (Mpa)

Section

85

85

Dimension

85

500

2000 500

1000

considered in parametric study. 40

500

Table 3.1: DetailFigure of cross sections 6.2: Cross-sections Table 2: Detail of cross-section

500

40 Steel position Rebar 500 500 500 500 y[m]the cross-section, column length

Steel sectioin

500 1000 1000 500 of relative slenderness and geometry of the b[m] h[m] 3000

500

30 (a) HEB 120 (b) HD 400x1086 0;0.02;0.05;0.1;0.3; 12 45 35;60;90 0;1 235;355;460 (e) Cross-section HSRCC4. 0.5;1;2;4;6 Figure 60 3.5: Detail of cross section of structural steel

Master thesis-MENG Try HSRCC5

500

3000

31

40

HBCol, the cross-section10of hybrid columns was divided into multi-fibres as show in [figure 0;0.02;0.05;0.1;0.3; HSRCC1 15 35;60;90 0;1 235;355;460 500 31 31 0.5;1;2;4;6 260 260 20 cross-section contains 3.2], each element of the different stress (σ ); strain (εi) and second i 800

40

2000

can

HSRCC1 0.8 HE120B 0.26 ; 0only, ; -0.26 the whole 16Φ20 range of be deduced. For 0.25 columns subjected to 3compressive load

(c) HSRCC5

Cross-section HSRCC3. 1 2

2 HD 400x1086 (d)

Cross-section HSRCC5. 0.5 ; -0.5 128Φ20

31

possible relative slenderness is covered. The parametrical study is summarized in Table 3.2: Rebar positions

Table 6.1.

250

5. Multiple results of Kc,EF; Ksa,EF Section X Y X Y 5.1. Correction factor for flexural stiffness of concrete Kc

X

Y

X

Y

31

-0.34 0.08 -0.23 0.08 about 0.00 31the strong 0.08 0.23 In this study, -0.08 bending is 31considered to take place axis. This 260 260

In the comparison of K0.00 other parameters, selected of HSRCC1 -0.08 with concrete 800 an example -0.08 we -0.11 -0.08 0.11 0.34 strength c,EF with -0.34 situation HSCRR 1corresponds to the case where the extreme load produced by wind or fck=60 Mpa; fyk=355 Mpa and vary with length0.08 and eccentricity can see an example in0.00 [Table 3]. The results 0.08 -0.34 -0.11 as we0.08 0.11 0.34 (e) Cross-section HSRCC4. of Kc,EF calculated from-0.08 finite element an appropriate also proposed -0.23 are put into -0.08the graph, 0.00 besides, -0.08 0.23 equation0.08 0.34 in each graph in order to compare with the expression of Kc,EC proposed by Eurocode 2. Before starting parametric study, it Figure 6.2: Cross-sections considered parametric study. is also necessary to sort the cases, which have large KSM, because engineerin would prefer to change the cross-section Section X Y X Y X Y X Y rather than increase material strength when this kind of situation happened.

Pisey Keo 133 -0.46 -0.46 0.96 -0.46 -0.84 -0.46 0.84 1.1 ≤ KEF ≤ 6 & 1.1 ≤ -0.96 KSM ≤ 6 of relative slenderness and geometry of the cross-section, length can -0.40 -0.96 -0.40 0.96 0.46 -0.84 the column 0.46 0.84

• •

𝜆≤2

-0.35 columns -0.96 subjected -0.35 0.96 -0.46 load -0.79 only, the -0.46 whole 0.79 range of be deduced. For to compressive -0.29 -0.96 0.96 0.46 -0.79 0.79 possible relative slenderness is-0.29 covered. The parametrical study0.46 is summarized in

Table 3: the

bending moment, flexural stiffness of concrete was not much reduce. In [figure 3.7 (b)], a ! general generalform formofofKKc cwith withNNEdEd/N /Npl,Rd was propose propose inin the the form formyy==ax ax !+ +c,c, and and the the correction correction pl,Rd was factor factorincrease increasewhile whileNNEdEd/N /Npl,Rd increase. Large Large axial axial force force with with small small end end bending bending moment, moment, pl,Rdincrease. flexural flexuralstiffness stiffnessofofconcrete concretedoes doesnot notreduce reduceasasfast fastas assmall small axial axial force force with with big big end end bending bending b) Kc,EF and b)(huge Kc,EF and φef φef This moment eccentricity). why in both graphs below have shown that K increases moment (huge eccentricity). This why in both graphs below have shown that K increases c,EF comparison From of Kthe c,EF c,EF with parameters [figure three specific cases the same material strength but different length From N the [figure 3.8 3.8 (a)],(a)], three specific cases withwith the same material strength but different length while /N/N while N alsoincreases. increases. EdEd pl,Rd pl,Rdalso Specific Case data of hybrid columns, value Kc was decrease along increasing φ. The increasing of hybrid columns, the the value of Kofc was decrease along the the increasing ofAll φ.ofThe increasing of of creep coefficient in concrete structure means to reduce elastic modulus (E=E =E /(1+φ)). c,eff cd creep coefficient in concrete structure means to reduce the the elastic modulus (Ec,eff /(1+φ)). It It cd !

! is logic to propose a general which provided a similar is logic to propose a general formform withwith y =y = !!!" which provided a similar formform as Kasc,EFKc,EF in in

Kc,EF Vs NEd/Npl,Rd

!!!"

[figure 3.8(a)]. plotted all data in the [figure 3.8(b)], it proved c is decrease [figure 3.8(a)]. We We alsoalso plotted all data in the [figure 3.8(b)], and and it proved that that Kc isKdecrease when φ is increase which is accordance to the first graph. Besides that there is a little when φ is increase which is accordance to the first graph. Besides that there is a little bit bit different between those graphs, figure a specific among cases different between those twotwo graphs, the the firstfirst figure is a isspecific casecase among 836 836 cases and and therethere value φ 0.5; (0; 0.5; 1; but 2), but in the figure is only (0;as1)we as we are are fourfour value of φof(0; 1; 2), in the figure (b), (b), therethere is only two two casecase of φof(0;φ 1) already stated in [section of parametric study. Furthermore, the equation proposed already stated in [section 3.4]3.4] casecase of parametric study. Furthermore, the equation proposed alsoalso ! !

! a similar form as that Kc Eurocode in Eurocode !!!!=!! !. It .isItalso is also necessary to noted be noted givegive a similar form as that of Kofc in !! = necessary to be in in !!! !!!

[figure φ starting to 0.5), means Kc of c,EF(L=10m)>K c,EF(L=20m), [figure 3.8 3.8 (a)],(a)], φ starting fromfrom (0 to(00.5), Kc,EFK(L=10m)>K this this means that that Kc of c,EF(L=20m), short columns is less reduce slender columns, in contrast, φ>0.5, short columns is less reduce thanthan slender columns, but but in contrast, for for φ>0.5, thenthen K(L=20m)>K While φ was a small value, concrete of slender columns subjected c,EF(L=20m)>K c,EF(L=10m). Kc,EF While φ was a small value, concrete of slender columns subjected c,EF(L=10m). to crack faster short columns, it was opposite for increasing the increasing of creep coefficient. to crack faster thanthan short columns, but but it was opposite for the of creep coefficient. (a) (a)

(b) (b)

Figure3.7: 3.7:Graph Graphof ofKc(EF) Kc(EF) Vs Vs NNEEdd/N /Nppl ,lR, Rdd Figure

Kc,EF Vs phi

Master thesis-MENG Try Master thesis-MENG Try

21 21

columns length. For example Kc,EF (L=10m)>K c,EF (L=20m) for the same concrete strength. The columns length. For example Kc,EF (L=10m)>K c,EF (L=20m) for the same concrete strength. The figure on the right side was the result of all data of Kcfck with the equation provided figure on the right side was the result of all data of Kc with andfcktheand equation provided has a has a columns length. example Kc,EF (L=10m)>K (L=20m) for the same concrete strength. columns length. ForFor example Kc,EF (L=10m)>K (L=20m) for the same concrete strength. TheThe c,EF c,EF ! !" !!" similar form compare to ! = in Eurocode 2. ! similar form compare to ! = in Eurocode 2. figure on the right side was the result of all data of K with f and the equation provided ! c fck and ck the equation provided has has figure on the right side was the!"result!"of all data of Kc with a a !

!" similar form compare !! =!!" in in Eurocode similar form compare to !to! = Eurocode 2. 2. !" !"

(a) (a)

Kc,EF Vs fck



(b) (b)

Figure Kc φVs Figure 3.8:3.8: Kc Vs ef φef



Kc,EF c) c) Kc,EF andand fck fck In [figure 3.9 3.9 (a)],(a)], a specific casecase of Kofc,EFK,c,EF in ,which Kc,EFKc,EF represented in axis ordinate and and axisaxis In [figure a specific in which represented in axis ordinate abscissa waswas fck with three different columns length has has beenbeen plotted. As we see, see, therethere are are abscissa fck with three different columns length plotted. As can we can three kinds of concrete strength (35,(35, 60 and 90 Mpa) and and Kc always increase when fck was three kinds of concrete strength 60 and 90 Mpa) Kc always increase when fck was increased. TheThe increasing of concrete strength willwill helphelp to reduce the crack of concrete under the the increased. increasing of concrete strength to reduce the crack of concrete under huge endend bending moment waswas applied. BothBoth [figure 3.9(a)&(b)] havehave proved Kc and fck infckthe huge bending moment applied. [figure 3.9(a)&(b)] proved Kc and in the ! !! !

form of yof=y = , which is anis increase function. Besides, in [figure 3.9(a)], a specific casecase withwith form , which an increase function. Besides, in [figure 3.9(a)], a specific ! !

three different length of column has has beenbeen introduced, the the increasing of Kofc was opposite to the three different length of column introduced, increasing Kc was opposite to the (a)

(b)

(a)

Master thesis-MENG Try Master thesis-MENG Try

(b)

22 FigureFigure 3.9: 22 Kc VsKc f c k Vs f c k 3.9: (a) (a)

(b) (b)

Kc,EF Vs Lamda

Figure Figure 3.9:3.9: Kc Kc Vs Vs fck fck

(a)

(a)

(b)

(b)

FigureFigure 3.10: 3.10: Kc VsKc Lamda Vs Lamda

In this paper, four comparisons between Kc,EF with various parameters (φ; fck; λ; NEd/Npl,Rd) as the same variables of d) Kd) c,EF and λ (b) (b) Kc,EF and λ(a) (a) Kc,EC shown in Eurocode 2. Besides that acolumns general was given parameter for both pair of graph Slenderness ratio hybrid has expression also played an important role to inrole Keach c,EF value. When the Slenderness of ratio of hybrid columns has also played an important in Kc,EF value. When the Figure 3.10: Kc Vs Lamda Figure 3.10: Kc Vs Lamda length of the column is increase, it means that the failure of the column is mostly caused by and compare to that of Eurocode 2. K be used to modify formulating Kc,new . c,EC will length of the column is increase, it means that thefor failure of the column is mostly caused by d) and Kc,EF and λ λ axial force will also reduce then the bending moment was diminished d) Kc,EF and instability the ultimate instability and the ultimate axial force will also reduce then the bending moment was diminished Slenderness ratio of hybrid columns hasthe also played an important role in Kinc,EF value. When Slenderness ratio of hybrid columns has also played an important role Kc,EF value. When compare to the short column and with same eccentricity. At the critical section, with thethe the compare to the short column and with the same eccentricity. At the critical section, with the length of the column istheincrease, it means thatthat the the failure of the column mostly caused by by length of the column isless increase, it moment means failure of ofthe is mostly caused same material strength, bending give bigger value K column . Inis [figure3.10 (a)], same material strength, the less bending moment give bigger valuec,EFof Kc,EF. In [figure3.10 (a)], instability andand the the ultimate axial force willwill alsoalso reduce thenthen the the bending moment waswas diminished instability ultimate axial force reduce bending moment diminished Master thesis-MENG Try 23 same compare to the short column andand withwith the the same eccentricity. At At the the critical section, withwith the the compare to the short column eccentricity. critical section,

5.2. Correction factor for equivalence flexural stiffness of structural steel and reinforced rebar (Ksa,EF) The comparison of Ksa,EF with other parameters will be plotted into two graphs as the previous section of Kc,EF and for the specific case, all curves shared the same material strength where fck=60 Mpa and fyk=355 Mpa. An appropriate equation for each graph will be given. Furthermore, four parameters (φ; fck; λ; NEd/Npl,Rd) will be plotted to study in this paper for generating new expression of Ksa,new.

Ksa,EF Vs phi

Ksa,EF Vs fyk

Ksa,EF Vs NEd/Npl,Rd

Table 4: the comparison of Ksa,EF with parameters Specific Case

All data

Ksa,EF Vs Lamda

kA

= ks =

COLUMN DESIGN

'2:.(/col/fcol) '2:.(a/beam/fbeam)

281

2 X 900 X 106/3.0 X 103 2 X 1 X 3125 X 106/4.0 X 103

= 0.38 fJ = 0.66, therefore Effective column height € 0 = 0.66 X €col = 0.66

From figure 9.3 take

X 3.0

= 1. 98 m

Slenderness Ratio ).

6. Failure mode of hybrid columns

3 . f . . /(/col) /12) h R a d ms o gyratiOn, z of = \jthe Aco, = \j/(bh bh 3 .46 = 86.6mm The failure mode columns is quite =important UMN DESIGN 281

for engineer to verify Slenderness ratio).the = €columns x 103 /86.6Related = 22.9 < 25 0 /i = 1.98 strength. 2 X 900 X 106/3.0 X 103 tothe member structure such as the columns under column isofshort. 2 Hence X1X 3125 X 106/4.0 X 103 axial =force with the (Nsd/(Acfcd) 1280 combine X 103/(400 X 300 Xbending 25/1.5) = moment, 0.64 thus 15/Y0.64 18.75 failure < 25 which is the greater value.) of column could be caused from cross-section

=

herefore

resistance buckling. = 0.66 X €col = 0.66 Xor3.0 = 1. 98 mThe end moment on a slender (4) Failure Modes column caused it to deflect sideways and in order to

Short columns usually fail by crushing but a slender column is liable to more The bending momenton(N.e further add). The fail byadd buckling. end moments a slender column cause it to deflectadditional sideways and thus bring into play an additional moment Neadd as 3 (/col) /12) bending moment also causes further Aco, illustrated = \j bh = 3 .49.5. 86.6mm in figure moment Neadd causes a further lateral 6 = The the(N) axial loada critical )value exceeds a deflection and if the axialif load exceeds this deflection, e shown thedeflection influence ofand parameters to the value of(NKEd sa,EF. In the next 3 /86.6moment the additional become self-propagating until the column = €0 and /i = 1.98 x 10 = 22.9 < 25 critical value of of this deflection, then the that additional formulate the new expression Ksa,EF with the parameters we have load for a pin-ended strut as buckles. Euler derived the critical moment until the on that was bending proposed above, will become be used toself-propagating combine by using the form of

/(bh

h

equation of column NEd/N=pl,Rd plus fykand that also take to influence Ksa,EF. X 300 X 25/1.5) 0.64 thus 15/Y0.64 = part buckle concrete cracks. value.)

Figure 5:

e columns is quite important for engineer toMom12nt verify Mthe columns strength. ushing but a slender column is liable to escribed in [section 3.3.1], related to member of structure such as the ments on a slender column cause it to rce combine with bending moment, the failure of column could be caused into play an additional moment Neadd as istance orNeadd buckling. Thea end moment on a slender column cause it to causes further lateral moment in exceeds order to aadd morevalue bending moment !!!"" . The further additional N) critical this deflection, come further self-propagating untilif the column causes deflection and the axial load (NEd) exceeds a critical value al load for a pin-ended strut as the additional bending moment become self-propagating until the column N acks. Figure 9.5

Mom12nt

Slender column with lateral deflection

M

N

column with lateral deflection

Figure 3.16: Hybrid column failure mode

example of a typical hybrid column of cross-section HSRCC1, where fck and length vary from (5;10;15;20;30m), was presented. The ratio of

In [fig. 5], an example of a typical hybrid column of cross-section HSRCC1, where fck=35 Mpa; fyk=235 Mpa and length vary from (5;10;15;20;30m), was presented. The ratio of Ncr/Nultimate determined the failure mode of the column. From the graph, it is noted that when λ is smaller than 40, the ratio of Ncr/Nultimate is bigger than 4 and this means that the apply load would cause the column to crash before it was buckled. In contrast, when the value of λ increase from 85, then the ratio of Ncr/Nultimate was almost constant and equal to one, at this stage our column is probably fail by buckling failure (instability) instead of crashing concrete. The detail of parametric study of Kc,EF and Ksa,EF with various parameters such as geometry and material strength were put into FEA, after finishing FEA for 1080 cases, all data was stored into matlab file and the comparisons between each correction factor (Kc,EF; Ksa,EF) to each parameter were presented by graph in [Table 3&4]. Each comparison has two kinds of graphs, first one is corresponding to a specific case among 836 cases and another one is to plot all data together. Furthermore, an appropriate equation also proposed to each graph. By using the proposed equations in each comparison, the combine of these equations is also the objective of this research, a new proposal of Kc and Ksa for calculating equivalence flexural stiffness of hybrid columns will be presented in the next section.

7. The new expression of correction factor Kc,new The new expression of Kc,new will be based on Kc,EC, there are three parts in the expression of Kc,EC.

•

The expression of Kc,EC

K !,!" =

1 ×K! ×K ! ; K! = 1 + φ!"

K ! = min 0.2,

!! !"#

; n=

f!" ; 20

!!" !!",!"

By using the equations given in the previous section [Table 3&4], there are four variables and equations were proposed corresponding to each part of Kc,EC in Eurocode 2, these equations will be combined to give a new expression of Kc,new for hybrid columns: •

The new expression of Kc,new K !,!"# =

Figure 6: Kc,EF Vs n;λ In [Fig. 6], it is a three variable graph, the idea is to formulate the formulation of K2 to take part into Kc,new based on the form proposed in EC2.

f!" K! = 20

1 ×K! ×K ! ≤ 1 ; 1.55 + φ!"

!.!

; K! =

370n! + λ N!" ; n= 395 N!",!"

Table 5: Equality line and expression of R Eurocode 2

New expression

R-square (percentage of successful fit) fit) • R-square (percentage of successful SSE SSE R !"#$%& = 1 −= 1 −= 88% R !"#$%& SST SST = 88% In twoIngraphs below,below, the ratio wasRgiven to Kc,EC (Eurocode) and the new expression Kc,newK , in two graphs the Rratio was given to K c,EC (Eurocode) and the new expression c,new, in whichwhich from the data in the in table the expression of Kc,new a better result than from thegiven data given the below, table below, the expression of Kprovide provide a better result than c,new Kc,EC for comparing with K . c,EF Kc,EF. Kc,EC for comparing with

Equality line

•

R=Kc,EFR=K /Kc,EC c,EF/Kc,EC

R=Kc,EFR=K /Kc,new c,EF/Kc,new Min(R)Min(R) = 0.63 = 0.63 Max(R)Max(R) = 1.99 = 1.99 Mean =Mean 1.07 = 1.07 Stdv. =Stdv. 0.22 = 0.22

Expression R

Min(R)Min(R) = 0.7 = 0.7 Max(R)Max(R) = 12.58= 12.58 Mean =Mean 2.4 = 2.4 Stdv. =Stdv. 1.86 = 1.86

FigureFigure 4.4: Statistic R = KRc , E=F /K 4.4: Statistic K cc,,En Fe w/K c , n e w

FigureFigure 4.5: Statistic R = KRc , E=F /K 4.5: Statistic K cc,,En Fe w/K c , n e w

With high of successful fit from fittingfitting of matlab, Kc,newK , ac,new nonlinear equation, With percentage high percentage of successful fit curve from curve of matlab, , a nonlinear equation, was considered as the most acceptable expression for fitting the data, furthermore, the was considered as the most acceptable expression for fitting the data, furthermore,10the 10 percentile value for the development of design equation was shown as below:

After modifying Kc,EC for generating new expression of Kc,new, in [Table 5], the comparisons of Kc,EC and Kc,new with Kc,EF were presented in the form of equlity line (45 degree), if both values approximate to each other, then they should stuck on the proposed equality line. It is obviously noted that Kc,EC is more conservative compare to Kc,EF. Besides, it is a significantly different between Kc,EC and Kc,new after observing two graphs above, the new equation of Kc,new is more approximate to Kc,EF compare to Kc,EC. e (percentage of successful fit) To prove the accuracy of Kc,new to Kc,EF, a statistic SSE analysis also explained [Table 5], the R !"#$%& = 1 −in the= 88% SST expression of ratio R was proposed:

8. The new expression of correction factor Ksa,new In the previous research paper of P. Keo [3], Ksa,pisey, an equivalence correction factor for both structural steel and reinforcing rebar, was based on two variables which are slenderness ratio and creep coefficient : 0.45λ!.!"# K !",!"#$% = 1 + 1.433φexp (−0.027λ)

In this paper, by considering NEd/Npl,Rd plus slenderness ratio; creep coefficient and fyk as the variables to formulate the new expression of Ksa,new. elow, the ratio R was!given to Kc,EC (Eurocode) and the new expression Kc,new, in !,!" In [Table 4], the appropriate equations correspond to • R= > 1 ∶ (464 cases = 55.5%) !!,!"# data given in the table below, the expression of Kc,new provide a better result than each parameters above were already introduced, we • . 0.9 < R < 1.1 ∶ (352 cases = 42%) ring with Kc,EF will renew the correction factor Ksa,pisey and combine R=Kc,EF/Kc,EC R=Kc,EF/Kc,new these equations to define Ksa,new : Min(R) = 0.7 Max(R) = 12.58 Mean = 2.4 Stdv. = 1.86

Min(R) = 0.63 Max(R) = 1.99 Mean = 1.07 Stdv. = 0.22

K !",!"# = K ! 2K! !.!" − K! ≤ 1; f!" !.! 314 K! = ; 1.2 + 0.6φexp (−0.02λ)

Table 6: 3D plot, equality line and histogram of R

.4: Statistic R = K c , E F /K c , n e w

Figure 4.5: Statistic R = K c , E F /K c , n e w

Equality line

entage of successful fit from curve fitting of matlab, Kc,new, a nonlinear equation, as the most acceptable expression for fitting the data, furthermore, the 10 for the development of design equation was shown as below: 352 0.9 < R < 1.1 case = 0.42 836

expression of correction factor Ksa,new

research of P.Keo, there are two versions of correction factor were proposed. The as Ka,pisey which was introduced in chapter 2 [section 2.2.3]. The second was quivalence correction factor for both structural steel and reinforcing rebar, the Ksa,pisey was based on two variable which are slenderness ratio and creep this master report, by considering NEd/Npl,Rd plus slenderness ratio; creep fyk as the variable to formulate the new expression of Ksa,new. In [section3.6.2],

K! =

N!" N!",!"

Min(R) = 0.526 Max(R) = 1.55 Max(R) = 1.55 Mean = Mean 1.04 = 1.04 Stdv. Stdv. = 0.187 = 0.187

Expression R

Min(R) = 0.56 Max(R) = 1.4 Max(R) = 1.4 Mean =Mean 0.99 = 0.99 Stdv. = 0.0957 Stdv. = 0.0957

Besides the graphs above, in order to assure the correction and acceptable of Ksa,new i Figure 4.11: R=K s a , E F /K s a , P i s e y matlab was based on the sum of square Figure 4.11: R=K s a , E F /K sdue a , P i s eto y error (SSE) and R

Figure 4.10: R=K s a , E F /K s a , n e w 4.8] curve Figure 4.10: , theR=K s a , Efitting F /K s a , n ein w

•

SSE (sum of square due to error)

high percentage offirstly successful fit, K , a nonlinear equation, was theorder most to formulate In the fist rowWith of [Table 6], we are finding the relation between K with λ; 4.75 φ and fas yk in SSEconsidered = With high percentage of successful sa,new fit, Ksa,new, a nonlinearsa,EF equation, was considered as the most acceptable expression for fitting the data, furthermore, the 10 percentile value for the expression of K3 based on the form of K it has shown in sa,pisey by using the curve fitting provided in matlab. As the • for R-square of successfulthe fit) 10 percentile value acceptable expression fitting(percentage the data, furthermore, for the development of design equation was shown below: [Table 4], the influence of N /N also played an important role for changing the results of K , for this reason, Ed pl,Rd sa,EF SSE development of design equation was shown below: R = 1 − = 72% NEd/Npl,Rd was considered to take part into the expression of Ksa,new !"#$%& in order to better SST estimate the equivalence 607 flexural stiffness of structural steel and reinforcing formulating 0.9 < R < rebar. 1.1 Aftercase 607= 0.73 Ksa,new, on the second row of [Table 6], 836 4.4. Verifying flexural stiffness of flexural stiffness calculated by (Kc,new & 0.9FE < and R < 1.1 case = 0.73 the comparison between of K4.10] with equality lineRwere sinceboth Ksa,pisey sa,new and sa,pisey 836 In the [figure and K[figure 4.11], the ratio was proposed, proposed for Ksa,EF and K Ksa,new) new expressions contained only two variables (φ;which λ), forthetheexpression same length and creep coefficient of hybrid column, K was a sa,pisey R was Ksa,EF divide by Ksa,new or Ksa,Pisey, and axis orthodontics rep As described in the chapter 3 [section 3.2], the flexural stiffness corresponding to each element Master thesis-MENG Try 34 constant and did not change withthe thefrequency strength of material, in which the data was arranged as a vertical line with in percentage. 34 For Ksa,new, there are 378 of conservative cases correspondi of hybrid columns cross-section was calculated from each fibre of the cross-section and at the different values ofMaster thesis-MENG Try Ksa,EF . K 378

!",!" most critical point. The sum of (EI)i,sec was the result of R (EI) is to =eff,EF. The idea > 1 of this chapter = 45.2% K 836 !",!"# develop an expression that would allow to calculate (EI)eff without doing finite element analysis. In the third row of [Table 6], the histogram of expression of ratio R corresponding to Ksa,pisey and Ksa,new were Withreduce both while ratio the R, load a comparison Ksa,new Ksa,Pisey factor with is Ksa,EF based on mean v Flexural stiffness is always is apply andofthat meansand a reduction presented, the conservative cases were counted for R>1. With both ratio R, a comparison of Ksa,new and Ksa,pisey needed to apply, moreover, Kc,new deviation and Ksa,new was played a roleinlike factor in to K obtain standard given thereduction table below, in order which sa,new is more accuracy than based on mean value and standard deviation was given in the table below, in which Ksa,new is more accuracy than (EI)eff,new. EC2 has also introduced correction factor (K and Ks =1)of which when compareits to K Ksa,new already was smaller than that of Ksa,P sa,EF, and because cof the Stdv. Ksa,Pisey when compare with Ksa,EF, 2and because of the of Kadapted than that of columns, Ksa,Pisey. sa,new was summarized in chapter [section 2.1.2], butStdv. it only to smaller reinforced concrete besides that, EC4 give a fix constant corresponding to concrete and steel profile and the R=Ksa,EF/Ksa,new R=Ksa,EF/Ksa,Pisey constants were briefly introduced in [section 2.1.3]. Compare to Eurocode, flexural stiffness of Min(R) = 0.56 Min(R) = 0.526 !!",!" columns wascases calibrated by Kc,new and Ksa,new instead of multiplying with aMax(R) constant as • 𝑅 =hybrid >1 : (378 = 45.2%) Max(R) = 1.4 = 1.55 !!",!"# describe in composite structure of EC4. Mean = 0.99 Mean = 1.04

•

1.1 > R > 0.9 : (607 cases = 73%) •

Eurocode 2:

Stdv. = 0.0957

Stdv. = 0.187

(EI)!"" = K ! E! I! + K ! (E! I! ) •

Eurocode 4:

9. Verifying (EI)EF and (EI)new(EI) calculated from Kc,new! I;! K +sa,new E! I! ) !"" = 0.45 E! I! + 0.9(E In order to verify acceptable • the Hybrid column:of the expressions of correction factors (Kc,new; Ksa,new), all cases of the equivalence flexural stiffness calculated form(EI) these correction factors used to = K !,!"# E! I! was + K !",!"# (E!assure I! + E!the I! ) accuracy of Kc,new and Ksa,new. !"",!"# (EI)eff,new = Kc,new(EcIc) + Ksa,new(EsIs + EaIa) Figure 7:

Figure 4.10: R=K s a , E F /K s a , n e w

Figure

Figure 4.11: R=K s a , E F /K s a , P

With high percentage of successful fit, Ksa,new, a nonlinear equation, was considered as acceptable expression for fitting the data, furthermore, the 10 percentile value 4.12:development (EI) E F Vs (EI) Figure below: 4.13: R ofn edesign equation was shown w

After having calculate all cases of flexural stiffness of hybrid columns by using Kc,new and

607

In [Fig. 7], the value of secant flexural stiffness of hybrid columns obtained from finite element analysis was plotted versus with equivalence flexural stiffness calculated from the expression of correction factors. Two statistic values of curve fitting from matlab also proved the accuracy of (EI)eff,new to (EI)EF, in which SSE presented the error of the point to equality line and R-square was the percentage that the points fit with the line: •

w ! u! − u

!

Hybrid colum:

M!",!"!#$ = K !" α! M!",! ; K !" =

N!"

π = l

!

β ; N 1 − !" N!"

N!" !.! (EI)!"",!"# ; α! = 1 − N!"

The comparison of magnification factor between KEF with KSM before and after calibrate with 𝛼! has been shown in [Fig. 8], the limitation of KSM before calibrate got out of the condition (KSM)Max =15>6, which we have already mentioned before starting

SSE (sum of square due to error) SSE =

•

•

= 1.2

R-square (percentage of successful fit)

parametric study. After having calibrate with 𝛼! , KSM and KEF was again put into compare in [Fig. 9] and most of the data given a good form which parallel to Besides the fitting curve, in R [figure the histogram of ratio in The histogram of ratio with4.13], the frequency shown in R with the thefrequency equality shown line, especially for the value between 1 percentage was developed to assure the correction between (EI) , in which, we can percentage was also developed to assure the new and (EI) to EF 1.5. see that high frequency was given between 0.9 to 1.1. Moreover, considering the conservative correction between (EI)eff,new and (EI)EF, in which, cases, there are 424 cases which have ratio R>1, and by considering all cases of parametric study high frequency was given between 0.9 to 1.1. mean value of R equalled 1.0026 and standard deviation was only 0.086. Considering EI the conservative cases, there are 424 650 424 !" R= >1 = 51% 0.9 < R < 1.1 case = 0.78 cases which EI have > 1. 836 !"# ratio R 836 R !"#$%& = 1 −

SSE = 89.4% SST

R = 0.7255 method • R>1:moment calculated byMin(R) 4.5. Bending magnification Max(R) = 1.32 424 cases = 51% An explanation concerning the method of bending moment magnification was already Mean(R) = 1.002 • 0.91 and 537 (64.23%) casesproposed of 0.95